Partial fractions

Partial fractions split one algebraic fraction into a sum of simpler fractions, one for each factor of the denominator. For example, (7x − 2)/((x + 1)(x − 2)) = 3/(x + 1) + 4/(x − 2). You choose the right form, multiply through by the denominator, then find the constants by substituting values of x or by equating coefficients. Below are worked partial fractions questions and a set to try, with answers.

Which form to use

AQA A-level Maths covers denominators no more complicated than squared linear terms, with no more than three terms and constant or linear numerators, and integration using partial fractions that are linear in the denominator. University courses add irreducible quadratic factors, which take a top of the form Bx + C. If the top has a power of x as high as the bottom, divide first.

How to find the constants

1. Write the form with unknown constants A, B, C.
2. Multiply both sides by the denominator, to get an identity with no fractions.
3. Substitute values of x that make each bracket zero. Each one kills all but one term.
4. Equate coefficients (compare the x² terms, x terms or constants) for any constant left over.
5. Check by substituting another value of x, or by adding the fractions back together.

Worked partial fractions questions

Question 1: two linear factors

Express (7x − 2)/((x + 1)(x − 2)) in partial fractions.

1. Write (7x − 2)/((x + 1)(x − 2)) = A/(x + 1) + B/(x − 2).
2. Multiply by (x + 1)(x − 2): 7x − 2 = A(x − 2) + B(x + 1).
3. x = 2: 12 = 3B, so B = 4.
4. x = −1: −9 = −3A, so A = 3.
5. Answer: 3/(x + 1) + 4/(x − 2). Check with x = 0: left side −2/(−2) = 1; right side 3 − 2 = 1.

Question 2: three linear factors

Express (2x² + 8x − 6)/(x(x − 1)(x + 3)) in partial fractions.

1. Form: A/x + B/(x − 1) + C/(x + 3), so 2x² + 8x − 6 = A(x − 1)(x + 3) + Bx(x + 3) + Cx(x − 1).
2. x = 0: −6 = −3A, so A = 2.
3. x = 1: 4 = 4B, so B = 1.
4. x = −3: 18 − 24 − 6 = −12 = 12C, so C = −1.
5. Answer: 2/x + 1/(x − 1) − 1/(x + 3).

Question 3: a repeated factor

Express (x + 5)/((x + 1)(x − 1)²) in partial fractions.

1. Form: A/(x + 1) + B/(x − 1) + C/(x − 1)². Multiplying through: x + 5 = A(x − 1)² + B(x + 1)(x − 1) + C(x + 1).
2. x = 1: 6 = 2C, so C = 3.
3. x = −1: 4 = 4A, so A = 1.
4. No value of x picks out B on its own, so equate the x² coefficients: the left side has no x², so 0 = A + B, giving B = −1.
5. Answer: 1/(x + 1) − 1/(x − 1) + 3/(x − 1)².

Equating coefficients often gives a pair of simultaneous equations. Practise solving a pair in this question from the Algebra course.

Question 4: integrating with partial fractions

Using Question 2, find the exact value of the integral below.

1. Split it: ∫ from 2 to 4 of (2/x + 1/(x − 1) − 1/(x + 3)) dx.
2. Each part integrates to a logarithm: [2 ln x + ln(x − 1) − ln(x + 3)] from 2 to 4.
3. At x = 4: 2 ln 4 + ln 3 − ln 7. At x = 2: 2 ln 2 + ln 1 − ln 5.
4. Subtract and use log laws: 2 ln 2 + ln 3 − ln 7 + ln 5 = ln(4 × 3 × 5 ÷ 7) = ln(60/7), about 2.15.

The log rules guide covers combining logarithms like this. Check any split in the Partial Fractions Calculator, which shows every step.

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Partial Fractions Calculator

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Partial fractions questions to try

1. Express (5x + 7)/((x + 1)(x + 3)) in partial fractions.
2. Express 3x/((x − 2)(x + 1)) in partial fractions.
3. Express (2x + 3)/(x + 1)² in partial fractions.
4. Use your answer to question 1 to find the exact value of ∫ from 0 to 1 of (5x + 7)/((x + 1)(x + 3)) dx.

Answers

1. 5x + 7 = A(x + 3) + B(x + 1). x = −1 gives A = 1; x = −3 gives B = 4. 1/(x + 1) + 4/(x + 3).
2. 3x = A(x + 1) + B(x − 2). x = 2 gives A = 2; x = −1 gives B = 1. 2/(x − 2) + 1/(x + 1).
3. 2x + 3 = A(x + 1) + B. x = −1 gives B = 1; equating x coefficients gives A = 2. 2/(x + 1) + 1/(x + 1)².
4. [ln(x + 1) + 4 ln(x + 3)] from 0 to 1 = ln 2 + 4 ln 4 − 4 ln 3 = ln 2 + 4 ln(4/3), about 1.84.

Common mistakes

• Missing a term for a repeated factor. (x − a)² needs both A/(x − a) and B/(x − a)².
• Not dividing first. If the top's degree is at least the bottom's, do polynomial division before splitting.
• Sign errors when substituting. For (x + 3), substitute x = −3.
• Forgetting the modulus or the chain rule when integrating. ∫ 1/(2x + 1) dx = ½ ln|2x + 1| + c, not ln|2x + 1|.
• Not checking. Substitute a convenient x into both sides; they must agree.

Practise with a short timed paper

Partial fractions are in AQA A-level Maths (B10, and H6 for integration) and first-year university calculus. The daily A-level Maths: Algebra & Functions and University Calculus papers are short timed papers of 8 questions, with a new one every day. For the calculator itself, see the partial fraction decomposition calculator guide.

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